使用 1 到 N 范围内的数字计算对中不同的 XOR 值

给定一个数字N 。任务是使用从1到N的数字来计算任何可能对的不同 XOR 的数量。

例子：

Input: N = 3
Output: 4
Explanation: Following are the all possible pairs using elements from 1 to N inclusive.
1^1 = 0 
1^2 = 3 
1^3 = 2 
2^2 = 0 
2^3 = 1
3^3 = 0 
Therefore, there are 4 distinct possible XOR values.

Input: N = 2
Output: 2

方法：这个问题是基于观察的。请按照以下步骤解决给定的问题。

• 对于N = 1和N = 2 ，所需输出分别为1和2 。
• 现在对于N≥3 ，有两种可能的情况：
  • 如果N不是 2 的幂，那么总共会有't'个数字，其中t是2的幂，紧邻数字N 。它们从[0 到 t – 1]不等。因为假设N = 5或6或7在上述所有情况下，值都从[0, 1, 2, 3, 4, 5, 6, 7]变化。
  • 如果N是 2 的幂，那么除了数字本身之外，所有数字都是可能的。因为假设N = 4在二进制中它是“100” ，所以我们知道如果两个位不同，XOR 运算给出“ 1” ，否则给出0 。可能的值为[0, 1, 2, 3, 4 , 5, 6, 7] 。

下面是上述方法的实现。

C++

// C++ code to implement above approach
#include 
#include 
using namespace std;
 
int findDistinctXOR(int n)
{
 
    // Corner case
    if (n == 1) {
        cout << 1 << endl;
    }
    if (n == 2) {
        cout << 2 << endl;
    }
 
    long long x, next;
    x = log2(n);
 
    // if n is power of two
    if ((n & (n - 1)) == 0) {
 
        next = pow(2, x + 1);
        cout << next - 1 << endl;
    }
    else {
        next = pow(2, x + 1);
        cout << next << endl;
    }
}
 
// Driver Code
int main()
{
    int N = 3;
 
    findDistinctXOR(N);
    return 0;
}

Java

// Java code to implement above approach
//include 
import java.util.*;
class GFG
{
 
static void findDistinctXOR(int n)
{
 
    // Corner case
    if (n == 1) {
        System.out.print(1 +"\n");
    }
    if (n == 2) {
        System.out.print(2 +"\n");
    }
 
    long x, next;
    x = (long) Math.log(n);
 
    // if n is power of two
    if ((n & (n - 1)) == 0) {
 
        next = (long) Math.pow(2, x + 1);
        System.out.print(next - 1 +"\n");
    }
    else {
        next = (long) Math.pow(2, x + 1);
        System.out.print(next +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 3;
 
    findDistinctXOR(N);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
import math as Math
 
def findDistinctXOR(n):
 
    # Corner case
    if (n == 1):
        print(1)
    if (n == 2):
        print(2)
    x = None
    next = None
    x = Math.floor(Math.log2(n))
 
    # if n is power of two
    if ((n & (n - 1)) == 0):
        next = Math.pow(2, x + 1)
        print((next - 1))
    else:
        next = Math.pow(2, x + 1)
        print(int(next))
 
# Driver Code
N = 3
findDistinctXOR(N)
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code to implement above approach
using System;
 
class GFG{
 
static void findDistinctXOR(int n)
{
     
    // Corner case
    if (n == 1)
    {
        Console.WriteLine(1);
    }
    if (n == 2)
    {
        Console.WriteLine(2);
    }
 
    long x, next;
    x = (long)Math.Log(n, 2);
 
    // If n is power of two
    if ((n & (n - 1)) == 0)
    {
        next = (long)Math.Pow(2, x + 1);
        Console.WriteLine(next - 1);
    }
    else
    {
        next = (long)Math.Pow(2, x + 1);
        Console.WriteLine(next);
    }
}
 
// Driver Code
public static void Main()
{
    int N = 3;
     
    findDistinctXOR(N);
}
}
 
// This code is contributed by ukasp

Javascript

       // JavaScript code for the above approach
       function findDistinctXOR(n)
       {
        
           // Corner case
           if (n == 1) {
               document.write(1 + "<br>")
           }
           if (n == 2) {
               document.write(2 + "<br>")
           }
           let x, next;
           x = Math.floor(Math.log2(n));
            
           // if n is power of two
           if ((n & (n - 1)) == 0) {
               next = Math.pow(2, x + 1);
               document.write((next - 1) + "<br>")
           }
           else {
               next = Math.pow(2, x + 1);
               document.write(next + "<br>")
           }
       }
        
       // Driver Code
       let N = 3;
       findDistinctXOR(N);
        
 // This code is contributed by Potta Lokesh
   

输出

4

时间复杂度： O(1)
辅助空间： O(1)